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Given that the greatest common divisor of $a$, $b$ and $c$ is $d$. And 3 doesn't divide $\frac cd$, $\frac cd$ is an even integer, $\frac {c}{2d}$ is an odd integer. Find the greatest common divisor of $\frac {12\gcd(a,b)}{d}$ and $\frac cd$

I've been struggling with this question for a while that appeared on my elementary number theory course in college and I initially thought that I should follow and use the definition of the greatest common divisor ($d=ax+by$) but now it seems to me like i should try and find some relations about $ \frac {12\gcd(a,b)}{d}$ and $\frac cd $ but i'm not quite sure how to connect the dots.

My initial guess was that, it is trivial to show $ \frac {12\gcd(a,b)}{d}$ is a multiple of $12$ but from that I'm lost. I did initially thought that $\frac cd $ is in the forms $12k+2$, $12k+6$, $12k+10$,(by using the given properties of the fraction) which meant I'd have to find a general insight of the greatest common divisor of those forms and the multiple of $12$ or $12k$ form which I'm yet to be wise of.

bb_823
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Nicholas Gray
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  • Follows immediately by a few mechanical applications of Euclid's Lemma, namely $,(12(a,b)/d,:!c/d) = (\color{#c00}{12},c/d),,$ by $\underbrace{((a,b)/d,:!c/d)=1}_{\textstyle \left[:!((a,b),c)=d:!\right]\div d}$ & $ $ Euclid's Lemma. & gcd distrib. law, therefore finally we have $,(\color{#c00}{12},c/d) = (4,c/d) = 2^{\phantom{|^{|^|}}}!!!!!,,$ by $,(3,c/d)!=!1,$ and $,2^1,||,c/d,,$ by Euclid's Lemma. – Bill Dubuque Sep 24 '23 at 19:06

2 Answers2

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Hint: if $\text{gcd}(a, b, c) = 1$, the only prime numbers that can divide simultaneously $12 a, 12b, c$ are 2 and 3.

Gribouillis
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  • thank you for your hint and yeah I can see what u mean. but if we generalize the greatest common divisor to be d, how would that affect and also how am I to generalize the common factors of 12k with the noted forms I listed above. if u could help me just a little more I'd really appreciate it – Nicholas Gray Sep 24 '23 at 13:43
  • This is imprecise handwaving without further details. Please provide them. – Bill Dubuque Sep 24 '23 at 17:45
  • @BillDubuque It is not handwaving, it is a sufficient hint to solve the problem. Th problem can be readily reduced to the case where $gcd(a, b, c)=1$. Then the hint gives the solution. – Gribouillis Sep 24 '23 at 18:42
  • Similarly one could hint "use unique factorization" for most all elementary number theory exercises. But that's far from a proof. Without any details it is handwaving. The point of elementary number theory classes is to teach how to give rigorous proofs, not how to handwave them away. If you give a hint and someone asks you to give further details then you should do so (what we expect on this site). – Bill Dubuque Sep 24 '23 at 18:48
  • @BillDubuque It is perfectly rigorous, with my handwaving the gcd of $12 a, 12b$ and $c$ must have the form $2^k 3^\ell$ but since $c$ is even and not a multiple of 4, and also not a multiple of 3, the gcd must be $2$. – Gribouillis Sep 24 '23 at 19:28
  • To be rigorous you need to justify your many unjustified claims, e.g. by naming the theorems you are implicitly using. Please do so in the answer, not in the comments. – Bill Dubuque Sep 24 '23 at 19:36
  • @BillDubuque I purposedly gave a hint, not a full proof. Rigorous reasoning is not the same matter as giving a prettily formatted proof. I'm sorry we disagree on this. – Gribouillis Sep 24 '23 at 19:48
  • If you are unable to give the requested details then it would be best to delete the answer since it may mislead beginners as to what constitutes a proof. – Bill Dubuque Sep 24 '23 at 19:55
  • @BillDubuque As the author of this reply, I have my editorial freedom. 3 persons upvoted it, you downvoted it, that is your right. Why should I delete my answer because you don't like it? – Gribouillis Sep 24 '23 at 19:59
  • This is not the place to have an extended discussion on the pedagogical pitfalls of elementary number theory. I was hoping that you would do the right thing and give the details of the proof you had in mind. But since you refuse to do so there is no point in continuing. – Bill Dubuque Sep 24 '23 at 20:06
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Simplify the problem by writing $$\gcd(a,b)=de,\quad c=2df.$$ The hypothesis become $$\gcd(e,2f)=1\quad\text{and}\quad\gcd(f,6)=1.$$

Now, $\gcd(e,2f)=1\implies\gcd(e,f)=1,$ and $$\gcd(f,6)=1\implies\gcd(f,6e)=\gcd(f,e)$$

so that $$\gcd(12e,2f)=2\gcd(6e,f)=2.$$

Anne Bauval
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